Originally Posted by

**thekrown** I have points p, q, r,

p = (-4,-1,-1)

q = (-2,0,1)

r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)

vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0

6y - 5z +1 = 0

I think that's right, what do you think?